Determine divergence or convergence of $\sum_{n=1}^\infty \frac{\cos\frac{\pi}{4n}}{\sqrt[5]{2n^5-1}}$

Question

Can anyone help with proof of divergence of this series? $$\sum_{n=1}^\infty \frac{\cos\frac{\pi}{4n}}{\sqrt[5]{2n^5-1}}$$

I have tried proof it using comparison criterion, so $b_n =-\frac{1}{\sqrt[5]{2n^5-1}}< \frac{\cos\frac{\pi}{4n}}{\sqrt[5]{2n^5-1}} = a_n $ , $\sum {b_n}$ is divergence , hence $\sum{a_n}$ is also should be divergent. But how can we test divergence of $\sum {b_n}$ , would integral test help?

Answer 1

Use the fact that $$\frac{1}{\sqrt[5]{2n^5-1}}\ge \frac{1}{\sqrt[5]{2}\cdot n}$$